I would it true because they do alot of experimenting
The number of moles of nitrate dissolved in water if 6.15 mole of strontium nitrate is : 12.3 moles
<u>Given data :</u>
Moles of strontium nitrate = 6.15 moles
<h3>Ionization of Strontium Nitrate </h3>
When strontium nitrate ( 
) is placed in water it will dissociates into its ions because strontium nitrate is an ionic compound.
Next step : Represent the ionization of strontium nitrate
From the Stoichiometry of the chemical reaction
I mole of strontium nitrate = 2 moles of Nitrate ions
6.15 mole of strontium nitrate = 2 * 6.15
= 12.3 moles of Nitrate
Hence we can conclude that The number of moles of nitrate dissolved in water if 6.15 mole of strontium nitrate is : 12.3 moles.
Learn more about Stoichiometry of chemical reaction : brainly.com/question/27058367
To convert from L to ml, we move three spaces to the right and the decimal place moves three places.
Our task here is to convert 600 L to mL. we have to remember that to convert from L to ml, we have to multiply the value in L by 1000
In other words, the value 1000 is known as the conversion factor.
When we do this, we move the decimal place three places to the right to have 600000.00mL
For more about conversion factors, see:
brainly.com/question/8512113
<span>To find the percent composition of KCLO(2), calculate the total mass of the molecule: K has 39.0983 g/mol, Cl has 35.4532 g/mol, and O(2) is 31.99886 g/mol. This sums to 106.5501 g/mol. Next, find the percentages of each element: K would have (39.0983/106.5501) = 0.366948 or 36.6948%, Cl would have (35.4532/106.5501) = 0.332737 or 33.2737%, and O(2) would be (31.99886/106.5501) = 0.300318, or 30.0318%. Summing these three percentages gives 100.0003%, which is within the scope of a rounding error.</span>
